Problem: Solve for $x$ and $y$ using elimination. ${6x-3y = 12}$ ${3x+5y = 45}$
Answer: We can eliminate $x$ by adding the equations together when the $x$ coefficients have opposite signs. Multiply the bottom equation by $-2$ ${6x-3y = 12}$ $-6x-10y = -90$ Add the top and bottom equations together. $-13y = -78$ $\dfrac{-13y}{{-13}} = \dfrac{-78}{{-13}}$ ${y = 6}$ Now that you know ${y = 6}$ , plug it back into $\thinspace {6x-3y = 12}\thinspace$ to find $x$ ${6x - 3}{(6)}{= 12}$ $6x-18 = 12$ $6x-18{+18} = 12{+18}$ $6x = 30$ $\dfrac{6x}{{6}} = \dfrac{30}{{6}}$ ${x = 5}$ You can also plug ${y = 6}$ into $\thinspace {3x+5y = 45}\thinspace$ and get the same answer for $x$ : ${3x + 5}{(6)}{= 45}$ ${x = 5}$